Chapter 9, Problem 38A

(a)

Interpretation Introduction

Interpretation:

For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.

  $CO\left(g\right)+2H_2\left(g\right)\to C{H}_{3}OH\left(l\right)$

Concept Introduction:

The mass of the product can be found out by multiplying number of moles to the molar mass.

Answer to Problem 38A

The limiting reagent is CO, and the mass of CH₃OH produced is 0.01142 grams.

Explanation of Solution

  $CO\left(g\right)+2H_2\left(g\right)\to C{H}_{3}OH\left(l\right)$

Now, calculate the number of moles of hydrogen and CO;

  $\begin{array}{l}{\text{n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{2 g/mol}}{\text{=5×10}}^{\text{-3}}\text{mol }\hfill \\ {\text{n}}_{\text{CO}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{28 g/mol}}\text{= 3}{\text{.57×10}}^{\text{-4}}\text{mol }\hfill \end{array}$

From the molar ratio CO:H₂ = 1:2

0.000357 mol of CO requires 2×0.000357 mol of H₂.

  $\begin{array}{l}CO{\text{ : H}}_2\\ 0.000357\text{ 0}\text{.0007142}\end{array}$

Hydrogen is in excess, hence the limiting reagent is CO.

  $\begin{array}{l}{m}_{CH3OH}=n\times M=3.57x10^{-4}mol\times 32g/mol=0.0114gCH3OH\left(11.4mg\right)\hfill \\ \hfill \\ \hfill \end{array}$

Limiting Reactant: CO

(b)

Interpretation Introduction

Interpretation:

For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.

  $\text{2Al}\left(\text{s}\right){\text{ + 3I}}_{\text{2}}\left(\text{s}\right)\to {\text{2AlI}}_{\text{3}}\left(\text{s}\right)$

Concept Introduction:

The mass of the product can be found out by multiplying the number of moles to the molar mass.

Answer to Problem 38A

The limiting reagent is I_2,and the mass of AlI₃ produced is 0.0107 grams

.

Explanation of Solution

  $\text{2Al}\left(\text{s}\right){\text{ + 3I}}_{\text{2}}\left(\text{s}\right)\to {\text{2AlI}}_{\text{3}}\left(\text{s}\right)$

  $\begin{array}{l}{\text{n}}_{\text{Al}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{27 g/mol}}\text{= 3}{\text{.70×10}}^{\text{-4}}\text{mol= 3}{\text{.70×10}}^{\text{-4}}{\text{mol AlI}}_{\text{3}}\hfill \\ {\text{n}}_{{\text{I}}_{\text{2}}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{253}\text{.8 g/mol}}\text{=3}{\text{.94×10}}^{\text{-5}}\text{mol }\hfill \end{array}$

Molar ratio of Al:I₂

=2:3

0.00003940 moles of iodine molecule required $\frac{2}{3}\times 0.00003940$ mol of Al

Al :      I₂

0.00002627    0.00003940

Al is in excess and hence iodine is the limiting reagent.

Molar ration of Al to AlI₂ =1:1

Moles of AlI₃ = 0.0000263 mol

  $m_{Al{I}_3}=n\times M=2.63\times 10^{-5}mol\times 407.7g/mol=0.0107gAl{I}_3\left(10.7mg\right)$

Limiting Reactant: I₂

(c)

Interpretation Introduction

Interpretation:

For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.

  $Ca{\left(OH\right)}_2\left(aq\right)+2HBr\left(aq\right)\to CaB{r}_2\left(aq\right)+2H_{2}O\left(l\right)$

Concept Introduction:

The mass of the product can be found out by multiplying the number of moles to the molar mass.

Answer to Problem 38A

Limiting Reagent is HBr, and mass of CaBr₂ = 0.01235 grams, Mass of H₂O = 1.112 x 10-3 grams

Explanation of Solution

  $Ca{\left(OH\right)}_2\left(aq\right)+2HBr\left(aq\right)\to CaB{r}_2\left(aq\right)+2H_{2}O\left(l\right)$

  $\begin{array}{l}{\text{n}}_{\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{74 g/mol}}\text{=1}{\text{.35x10}}^{\text{-4}}\text{mol }\\ {\text{n}}_{\text{HBr}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{81 g/mol}}\text{= 1}{\text{.23×10}}^{\text{-4}}\text{mol}\end{array}$

Molar ratio of Calcium hydroxide: HBr=1:2

0.000123 mol of HBr required 1/2×0.0000123 moles of calcium hydroxide

HBr : Ca(OH)₂0.000123 0.0000618

Hence, calcium hydroxide is in excess, limiting reagent is HBr.

  $\begin{array}{l}{\text{moles of CaBr}}_2=\frac{1}{2}\times 0.000123=6.17x10^{-5}mol\\ m_{CaB{r}_2}=n\times M=6.17x10^{-5}mol\times 200g/mol=0.0123gCaB{r}_2\left(12.3mg\right)\\ m_{H_{2}O}=0.0000618mol\times 18g/mol=1.112\times {10}^{-3}g\end{array}$

Limiting Reactant: HBr

(d)

Interpretation Introduction

Interpretation:

For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.

  $Cr\left(s\right)+H_{3}P{O}_4\left(aq\right)\to CrP{O}_4\left(s\right)+3/2H_2\left(g\right)$

Concept Introduction:

The mass of the product can be found out by multiplying the number of moles to the molar mass.

Answer to Problem 38A

Limiting Reagent is H₂PO_4,and mass of CrPO₄ = 0.0151 grams, Mass of H₂ = 2.0621 x 10-4 grams.

Explanation of Solution

  $Cr\left(s\right)+H_{3}P{O}_4\left(aq\right)\to CrP{O}_4\left(s\right)+3/2H_2\left(g\right)$

  $\begin{array}{l}{\text{n}}_{\text{Cr}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{52 g/mol}}\text{= 1}{\text{.92×10}}^{\text{-4}}\text{mol= 1}{\text{.92×10}}^{\text{-4}}{\text{mol CrPO}}_{\text{4}}\\ {\text{n}}_{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=}\frac{{\text{10×10}}^{\text{-3}}\text{g}}{\text{98 g/mol}}\text{=1}{\text{.02×10}}^{\text{-4}}\text{mol}\end{array}$

Molar ratio of Cr:H₂PO₄=1:1

0.000102 mol of H₂PO₄required 0.000102 moles of Cr.

Hence, calcium hydroxide is in excess, limiting reagent is HBr.

  $\begin{array}{l}{m}_{CrP{O}_4}=n\times M=1.02x10^{-4}mol\times 147g/mol=0.0150gCrP{O}_4\left(15.0mg\right)\\ massof{H}_2=0.000102mol\times 2g/mol=2.04\times 10^{-4}g\end{array}$

Limiting Reactant: H₃PO₄